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If $v$ is a compactly supported distribution and $u$ is any distribution, define $⟨u\ast v,\varphi ⟩=⟨u_x,⟨v_y,\varphi (x+y)⟩⟩$. The inner function is smooth and supported in $\mathrm{supp}\varphi -\mathrm{supp}v$, hence is a test function for $u$. Finite-order estimates prove continuity, while the tensor product of distributions proves commutativity after inserting compact cutoffs. For every test function $\psi$, $(u\ast v)\ast \psi =u\ast (v\ast \psi )$; evaluating at zero against reflected test functions proves uniqueness. If both distributions have compact support, $\mathrm{supp}(u\ast v)\subset \mathrm{supp}u+\mathrm{supp}v$ and the convolution theorem gives $\widehat{u\ast v}=\hat{u}\hat{v}$. Without a compact factor or another support/growth condition, arbitrary distributional convolution is not generally defined.

The Fourier transform of a Gaussian and the convolution theorem give this multiplier with the angular-frequency convention $e^{-ix\cdot \xi }$. In cycles-per-length frequencies $k$, it is $e^{-2{\pi }^2{\sigma }^2|k{|}^2}$. Each coordinate has variance ${\sigma }^2$, and higher frequencies receive stronger attenuation.

Use complex-linear distribution pairings, without conjugation. Write $\mathcal{D}=C_c^{\infty }({\mathbb{R}}^n)$ for the space of test functions. The smoothing convolution with a test function isOn a compact set of $x$ values, all the translated test functions have support in one compact set. Continuity of the distribution therefore permits differentiation in $x$, giving ${\partial }_x^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$ for every multi-index. In particular this convolution is a smooth function, even if $u$ is not tempered.

For the first associativity identity, integration against $\psi$ and the distribution pairing can be interchanged: the integrand has a common compact support in $y$, depends smoothly on the integration variable, and satisfies the finite-order continuity estimate there. ConsequentlyThe common-support argument matters: a general distribution cannot be paired with arbitrary noncompact functions.

For convolution of distributions with a compactly supported factor, first take $v$ with compact support $K$ and defineThe inner pairing is interpreted using a cutoff function equal to one near $K$. It is smooth in $x$ and has support in $\mathrm{supp}\chi -K$. To see continuity, restrict $\chi$ to a fixed compact support $A$. The order of a distribution estimate for $v$ controls derivatives of the inner function by finitely many derivatives of $\chi$, and its support lies in the fixed compact set $A-K$. Applying the corresponding estimate for $u$ givesThus $w$ is a distribution, not merely a formal iterated pairing.

Choose an additional cutoff function in $x$ equal to one on a neighborhood of $A-K$. The resulting joint kernel is compactly supported in both variables, so the tensor product of distributions permits reversing the pairings. One justification is to approximate that smooth compact kernel, in all the required derivative seminorms, by finite sums of products of one-variable kernels; the two orders agree on such products and their continuity estimates pass to the limit. Reversing $x,y$ consequently gives $u\ast v=v\ast u$. If $u$ rather than $v$ has compact support, use the same construction with the roles reversed; pairing $u$ against a smooth function is then legitimate.

Evaluating the resulting smoothing convolution with a test function givesIf $v$ has compact support, $v\ast \phi$ is itself a test function; if $u$ has compact support, its action on the smooth inner convolution uses a cutoff. This explains the meaning of the formula in either case. It also proves uniqueness: $(w\ast \phi )(0)=⟨w,\phi (-\cdot )⟩$, and reflection runs through all test functions. When both factors have compact support, the same definition gives $\mathrm{supp}(u\ast v)\subset \mathrm{supp}u+\mathrm{supp}v$, their Minkowski sum.

The Schwartz space consists of smooth functions for which every seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }\varphi (x)|$ is finite. A tempered distribution is a continuous linear functional on this space. Fix the angular-frequency Fourier transform conventionThe Fourier transform isomorphism of the Schwartz space makes the dual definition continuous. For a compactly supported distribution, a fixed cutoff function $\chi =1$ near its support extends the action to smooth functions by $⟨u,g⟩=⟨u,\chi g⟩$. A finite-order estimate controls this by finitely many Schwartz space seminorms, so both compactly supported factors are tempered.

Their Fourier transform of a compactly supported distribution is the smooth function $\hat{u}(\lambda )=⟨u,e^{-i\lambda \cdot x}⟩$. Applying the compact-support convolution definition to the exponential givesThe convolution theorem has no extra factor with this normalization.

For the spherical surface measure convolution, put $k=|\lambda |$. Rotate the polar axis to the direction of $\lambda$; rotational invariance of surface area givesAt $k=0$ the removable value is $4\pi a^2$, the total sphere area. Hence $\widehat{u_a\ast u_b}(\lambda )=16{\pi }^{2}ab\sin (ak)\sin (bk)/k^2$.

For $r=|x|>0$, angular integration in Fourier inversion now givesThis conditional integral can be made rigorous by first inserting $e^{-\epsilon k}$ and then taking $\epsilon \downarrow 0$ in tempered distributions. The supplied sine identity gives an integral of $\pi /4$ when $|a-b|<r<a+b$, and zero off that interval. Thusas a regular distribution. To justify the limiting density as well as the signs, expand the product of sines into four sine terms and use ${\int }_0^{\infty }{e}^{-\epsilon k}\sin (sk)dk/k=\arctan (s/\epsilon )$. The four arctangents are uniformly bounded; the regularized inverse is bounded by a constant times $1/r$, which is a locally integrable function in three dimensions. Dominated convergence theorem therefore identifies the distributional limit with the displayed density.

Changing the two endpoint sphere values does not change the regular distribution; this includes the source's closed-interval representative. At a jump, symmetric Fourier inversion instead takes the half-value. If $a=b$, the $1/r$ singularity at the origin remains locally integrable and is not a point mass. As a normalization check,exactly the product of the original sphere areas.

For smooth rapidly decaying initial data, apply the Fourier transform in $x$ and differentiate under the integral. The Fourier transform of a derivative turns the wave equation intoSolving this constant-coefficient ordinary differential equation givesAt $k=0$ the quotient means its continuous limit $t$, so $\tilde{v}(0,t)=\tilde{f}(0)+t\tilde{g}(0)$.

Since $\cos kt=(e^{ikt}+e^{-ikt})/2$, the Translation property of the Fourier transform inverts the first term as $(f(x+t)+f(x-t))/2$. Also $K_t(x)=\frac{1}{2}{1}_{[-t,t]}(x)$ has Fourier transform $\sin kt/k$, including its limiting value at zero. The convolution theorem therefore inverts the second term asThis proves the D'Alembert formulaThe formula itself extends beyond the rapidly decaying class: $f\in C^2$ and $g\in C^1$ suffice for a classical solution, verified by differentiation and the two initial conditions. The decay assumption above justifies the ordinary transform derivation without distributional qualifications.

For $n\ge 1$, the convolution equals the length of the overlap of two intervals:It is continuous and compactly supported, with $\Vert g_n{\Vert }_{\infty }=2$. Inverse Fourier transform turns this convolution into a product, soThis function is bounded near zero and $O(x^{-2})$ at infinity, hence lies in $L^1$. Fourier inversion, or the convolution theorem applied first to the integrable interval indicators, gives ${\hat{h}}_n=g_n$.

Use the angular-frequency Fourier transformThenwith the continuous value $\hat{f}(0)=2A$.

The convolution $(f\ast f)(x)$ is $A^2$ times the length of the overlap of the intervals $[-1,1]$ and $[x-1,x+1]$. Thus

The convolution theorem states, for this convention, thatSince $g=(B/A^2)(f\ast f)$, it follows thatagain interpreted continuously at $k=0$, where $\hat{g}(0)=4B$.

Termwise differentiation of a Fourier series multiplies a mode by its wavenumber, soThe convolution theorem turns pointwise multiplication into discrete convolution:soThe assumed real analyticity justifies these termwise operations through rapid decay of the Fourier coefficients.

Extend the boundary data oddly to the whole real axis. The extended function is precisely the function $f$ from part (a). Taking the Fourier transform in $x$, the Laplace equation becomesDecay as $y\to \infty$ selectsPart (b), together with the convolution theorem, says that the inverse transform is convolution with the Poisson kernelFor $x,y>0$ this givesThe odd extension enforces $u(0,y)=0$, while the Poisson integral takes the prescribed boundary values at every continuity point and decays at infinity.

The transform factors as $s^{-2}(s+2{)}^{-2}$, whose inverse factors are $t$ and $t{e}^{-2t}$. The Laplace convolution theorem givesand direct integration producesconfirming part (b).

Because the centered normal distribution density $g_t$ is nonnegative and has integral one, Fubini's theorem and translation invariance giveThus $f\ast g_t\in L^1$.

Use the angular-frequency conventionThe convolution theorem and the Gaussian transform giveSince $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:This proves the Fourier inversion theorem for the convolution directly.

Now additionally suppose $\hat{f}\in L^1$. For every $x$, the dominated convergence theorem givesFourier inversion identifies the right-hand side with $f(x)$ for Lebesgue almost everywhere $x$. Therefore the Gaussian approximate identity satisfies

For the stated angular-frequency Fourier transform convention, the convolution theorem givesThe Fourier inversion theorem and the assumed absolute integrability therefore imply

Since $|-i|=1$ and the principal argument is $-\pi /2$,All logarithms are $-\pi i/2+2\pi in$, $n\in \mathbb{Z}$.

Take the spatial Fourier transform of the wave equation. The transformed initial-value problem isand hence

Part (c) and the convolution theorem turn the first term intoFor the second term, choose $p$ with $p^{\prime }=g$. Part (b) says $\tilde{g}=ik\tilde{p}$, soPart (c) now makes its inverse transformCombining the two terms gives the D'Alembert formula

Let $A=2V$ and $B=W-V$. Part (c) makes $A$ and $B$ independent identically distributed exponential variables of rate $\lambda$. The change of variables $(A,B)=(rs,(1-r)s)$ has Jacobian $s$, and integrating ${\lambda }^{2}s{e}^{-\lambda s}$ over $s>0$ gives density one for $0<r<1$. Therefore the uniform ratio of independent exponential variables gives

By the Fourier inversion theorem and Fubini's theorem,Thus multiplication of Fourier transforms corresponds to the convolution $h=f\ast g$, which is the convolution theorem in the convention used here.